The study of forces resulting from the impact of fluid jets and when fluids are diverted round pipe bends involves the application of newtons second law in the form of F = m.a. The forces are determined by calculating the change of momentum of the flowing fluids. In nature these forces manifest themselves in the form of wind forces, and the impact forces of the sea on the harbour walls. The operation of hydro-kinetic machines such as turbines depends on forces developed through changing the momentum of flowing fluids.

I have created a excelcalcs spreadsheet for convenient access to all of the equations found on this page . This is located at ExcelCalcs.com calculation Fluid Jets

α = jet angle (radian) |
u _{1} = initial velocity (m/s)u _{2} = final velocity velocity (m/s)P = Power (watts) Q = Volumetric Flow Rate (m ^{3}/s)θ _{1} vane inlet angle angle (radian)θ _{2} vane outlet angle angle (radian) |

Newtons Second Law can be stated as: The force acting on a body
in a fixed direction is equal to rate of increase of momentum of the body
in that direction. Force and momentum are vector quantities so the direction
is important. A fluid is essentially a collection of particles and the net force, in a fixed direction,
on a defined quantity of fluid equals the total rate of momentum of that fluid quantity
in that direction.

Consider a mass m which has an initial velocity u and is brought to rest. Its loss of momentum is m.u and if
it stopped in a time interval t then the rate of change of momentum is m.u /t.
The force F required to stop the moving mass is therefore F = m.u / t . Now if this is applied
to a jet of fluid with a mass flow rate ( m / t ) which is equivalent to the volumetric flow rate times the density ( Qρ )
the equivalent force on a flowing fluid is F = Qρ u. Also in accordance with Newtons
third law the resulting force of the fluid by a flowing fluid on its surroundings is (-F).
Newtons third law states that for every force there is an equal and opposite force.

The figure below illustrates this principle at two locations.

The fluid flowing
into the tank is brought to rest from a velocity u to zero velocity and the force on the jet
is F = Q.ρ. u. . The reaction force on the tank contents is -F.

The fluid flowing in the pipe in the horizontal direction is forced to change direction at the bend such that
its velocity in the orginal direction is zero. Therefore the force on the flowing
fluid is F = Q. ρ u. The reaction force on the pipe is -F in the horizontal
direction as shown.

In its simplest form, with steady flow conditions, the force on a fluid flow in a set direction is equal to its mass flow rate times by the change in velocity in the set direction. The fluid flow also exerts an equal and opposite reaction force as a result of this change in momentum.

**F = Qρ (u _{1} - u _{2}). ..( F and u are vector quantities)**

**The resultant force on a fluid in a particular direction is equal to the rate of increase of momentum
in that direction.**

__Jet force on a flat plate__

Considering only forces in a horizontal direction u _{1} = V and u _{2} = 0 therefore

F = QρV = ρAV ^{2}

__Jet force on a flat plate at an angle θ __

Considering only forces Normal to plate surface u _{1} = V sinθ and u _{2} = 0 therefore

F = QρV sin θ = ρAV ^{2} sin θ

when θ =90^{o} then F = ρAV ^{2} as above

__Jet force on an angled plate.__ (θ < 90 ^{o})

Considering only forces in a horizontal direction u _{1} = V and
u _{2} = V cos θ therefore

F = QρV (1 - cosθ ) = ρAV ^{2}(1 - cosθ )

__Jet Force on an angled plate __(θ > 90 ^{o})

Considering only forces in a horizontal direction
u _{1} = V and
u _{2} = V cos θ therefore

F = QρV (1 - cosθ ) = ρAV ^{2}(1 - cosθ )

__Jet Force on an angled plate __(θ = 180 ^{o})

u _{1} = V and
u _{2} = V cos θ therefore

F = QρV (1 - cos180 ^{o} ) = 2QρV = 2ρAV ^{2}

__Jet force on a moving flat plate__

Considering only forces in a horizontal direction

u _{1} = V and
u _{2} = V p

and let r = V _{p} / V therefore

F = Qρ( V - V _{p} ) = ρAV(V -V _{p})

and F = ρAV ^{2}( 1 - r

The power (P) generated by the force on the moving plate = P = F. V p

__Jet force on an angled moving plate__

Considering only forces in a horizontal direction

u _{1} = V and
u _{2} = V _{p } + (V - V _{p}) cos θ

and let r = V _{p} / V therefore

F = ρA V( V - V _{p} ) ( 1 - cosθ ) = ρA V ^{2} ( 1 - r ) ( 1 - cosθ )

The power (P) generated by the force on the moving plate P = F. V p

Note: The moving plate with an angle θ = 180^{o} is the typical of the rotating plate of the pelton wheel.

The ideal value for r resulting in the maximum power output is clearly 0,5

Additional notes can be found on webpages

Fluid Machines - Francis Wheel

Steam Turbines - Impulse blades

__Force on fixed Vane.__

In the x Direction: u _{1x} = V cos θ _{1} ,
u _{2x} = -V cos θ _{2}

F _{x} = QρV(cos θ _{1} + cos θ _{2} )
= ρAV ^{2}(cos θ _{1} + cos θ _{2} )

In the y Direction u _{1y} = V sin θ _{1}
u _{2y} = V sin θ _{2}

F _{y} = QρV(sin θ _{1} - sin θ _{2} )
= ρAV ^{2}(sin θ _{1} - sin θ _{2} )

__Force on Moving Vane.__

The notes below related to vanes as used in impulse turbines. These turbines
derive the mechanical energy mainly from the change in momentum as the fluid passes through the vanes. The conditions as shown when
the vectorial sum of V _{v} + V _{r1} = V_{1} results in smooth entry with efficient transfer of energy of the fluid to the vane.
When this does not occur there will be turbulent flow over the vane with significant losses.

In the x Direction

u _{1x} = V _{1} cos α

u _{2x} = V _{v} - V _{r2} cos θ _{2} ....
( V _{r2} = V _{r1} = V _{1} sin α /sin θ _{1} )

F _{x} = QρV _{1} (cos α + [sin α /sin θ _{1} ] cos θ _{2} - r > )

r = V _{v} / V _{1}

In the y Direction u _{1y} = V _{1} sin α

u _{2y} = V _{r2} sin θ _{2} ....
( V _{r2} = V _{r1} = V _{1} sin α /sin θ _{1} )

In the y direction
F _{y} = QρV _{1}sin α (1 - sin θ _{2} / sin θ _{1} )

If the vane is moving in the x direction the power developed by the vane P = F _{x}.V _{V}

__Force on Pipe Wall__

The notes below related to the force on a pipe wall resulting from the changes in fluid pressure and fluid momentum as the fluid flows round a pipe bend . Gravity and friction effects are not considered. The fluid is assumed to be flowing under steady state conditions.

In the x Direction

u _{1} = V _{1}

u _{2} = V _{2}

F _{x} = p_{1}.A_{1} - p_{2}.A_{2} + ρ (A_{1}V_{1}^{2} - A_{2}V_{2})^{2}

In the x Direction u _{1x} = V _{1} u _{2x} = V _{2}cos θF _{x} = p_{1}.A_{1} - p_{2}.A_{2}cos θ + ρ (A_{1}V_{1}^{2} - A_{2}V_{2}^{2} cos θ ) |
In the y Direction u _{1y} = 0 u _{2y} = V _{2}sin θF _{y} = - p_{2}.A_{2}sin θ - ρA_{2}V_{2}^{2} sin θ |

The resultant reaction force on the pipe = F_{r} =√ (F_{x}^{2} + F_{x}^{2} )

The angle α of the resultant force to the x axis = tan^{-1} ((F_{y} /(F_{x})

- The Momentum equation...Useful lecture notes
- Eulers equation for fluid flow .. One page derivation from basic principles
- Turbines .. Notes on turbines explaining the difference between impulse and reaction types